Additioal Brach ad Boud Algorithms 0-1 Mixed-Iteger Liear Programmig The brach ad boud algorithm described i the previous sectios ca be used to solve virtually all optimizatio problems cotaiig iteger variables, but problem classes will differ i the implemetatio of the subrouties Brach, Approximate, ad Variable Fixig. We ow discuss some of the issues that arise whe tryig to solve a 0-1 MILP. The problem has the followig form Maximize c 1 x + c 2 y subject to A 1 x + A 2 y b x j 0, j = 1,, 1 y j = 0 or 1, j = 1,, 2 where x is a 1 -dimesioal vector of real variables, y is a 2 - dimesioal vector of biary variables, ad = 1 + 2. The umber of structural costraits is m, ad the parameters c 1, c 2, A 1, A 2 ad b are appropriately sized arrays. Boud The most obvious relaxatio of this problem is obtaied by removig the itegrality requiremets o the biary variables. The result is a bouded variable liear program with the same format as above except the itegrality requiremets are replaced by 0 y j 1, j = 1,, 2 Assume at some poit i the eumeratio process that the sets S + k, S k ad S 0 k are available. The liear programmig relaxatio of the problem is z k UB = Maximize c 1 x + Â c 2j y j + Â c 2j jœs 0 k subject to A 1 x + A 2 y b jœs + k x j 0, j = 1,, 1
0 y j 1, j ΠS 0 k y j = 1, j ΠS + k y j = 0, j ΠS k where the third term i the objective fuctio is a costat, ad the last two costraits fix a subset of the biary variables to either 0 or 1. The solutio to this LP provides a upper boud to the set of solutios associated with ode k. If the problem is ifeasible, the ode ca be fathomed. Note that we ca o loger use Eq. (4) to reduce z k UB by its fractioal part ad thus obtai a more powerful boud. Because some of the variables are allowed to be oiteger, the optimal objective value may be fractioal. Approximate Obtaiig a feasible iteger solutio from the relaxed solutio is ot, i geeral, simple sice roudig will rarely produce the desired result. Of course, if the relaxed solutio has all iteger values for y, it is also feasible to the origial problem. I this case, it is cosidered as a replacemet for the icumbet y B i the Update subroutie. I ay case, whe the relaxed solutio satisfies the itegrality requiremet the ode is fathomed ad backtrackig occurs. Example 3 (Facility Locatio Problem) To illustrate this brach ad boud procedure for a 0-1 MILP we cosider the facility locatio problem discussed i Sectio 7.4. Figure 8 gives the uit costs of trasportig a commodity from each of three proposed warehouse locatios to five differet customer sites. The demad for the commodity is give at the bottom of the matrix. Customer Warehouse 1 2 3 4 5 1 15 15 16 11 11 2 13 11 15 9 6 3 8 12 11 7 8 Demad 5 10 15 5 10
Figure 8. Data for facility locatio problem We must decide which of the three locatios should have a warehouse. The capacity of a warehouse, if built, is 30. The cost of buildig a warehouse at locatio 1, 2 or 3 is 200, 300 ad 200, respectively. The problem is a simplificatio of the oe preseted i Sectio 7.4. We use the model that defies the variables as a proportio of demad rather tha as the amout shipped. The correspodig LP relaxatio has a feasible regio that is closer to the MILP feasible regio ad provides better bouds for the fathomig test. Recall that the decisio variables are x ij = fractio of jth customer s demad met from warehouse i y i = 1 if warehouse is built at locatio i; 0 otherwise ad the model is: Miimize z = m  i =1  d j c ij x ij +  f i y i j =1 m i =1 m x ij i =1 (All demad must be met)  = 1, j = 1,, (Warehouse capacity limits)  d j x ij u i y i, i = 1,,m j =1 (Noegativity) x ij 0, i = 1,,m; j = 1,, (Itegrality) y i = 0 or 1, i = 1,,m x ij y i, i = 1,,m; j = 1,, The last set of m costraits is redudat, but its iclusio ca icrease the computatioal efficiecy of the B&B algorithm sigificatly. The idea is to explicitly limit the use of ay shippig lik whose correspodig warehouse is ot opeed. The costraits force the values of the y i variables to assume the maximum value of the amout shipped (that is, max j x ij ) from warehouse i. This produces a tighter LP feasible regio ad hece better bouds. To form the liear programmig relaxatio, the itegrality costraits are replaced by simple bouds 0 y i 1. The search tree resultig from the brach ad boud computatios is illustrated i Fig. 9. Table 9 provides the details for each iteratio. The vector y LB gives the
z = 910.0 F relaxed solutio for the 0-1 variables with z LB the objective value. Sice we are miimizig rather tha maximizig, the LP returs a lower boud rather tha a upper boud as was the case up util ow. This meas that we must reverse the sese of the iequality i coditio (3). With this modificatio, the appropriate test for fathomig ode k becomes z k LB z B (5) The separatio variable is idetified i the colum labeled s ad was chose to be the y i closest to 0.5 amogst all fractioal y i ; that is, y s = mi i { y i 0.5 }. Whe the value of the fractio was exactly 0.5, the y s = 1 ode was explored first. Because we are solvig a mixed-iteger program, coditio (4) ca o loger be use to roud the LP objective value. z = 860.0 LB +2 0 z = 790.0 LB 1 4 +3-3 +1-1 z = 932.5 LB z = 857.5 2 3 5 LB (0, 1, 1) Feasible z = 885.0 B +3-2 -3 6 7 (1, 0, 1) Feasible Ifeasible Figure 9. Search tree for warehouse locatio problem z = 792.0 LB 8 Ifeasible The first feasible solutio is ucovered at ode 2 yieldig a objective value z F = 910. The algorithm the backtracks to ode 3 where the LP objective value is z LB = 932.5. Coditio (5) allows us to fathom this ode so we backtrack to ode 4. A secod feasible solutio is foud at ode 6 which turs out to be the optimum. This is cofirmed after examiig two more odes. A fial poit about the example is that from the data we see that the total demad is 45 ad the capacity of each warehouse is 30. Cosequetly, all solutios that specify oly oe warehouse ca be fathomed immediately without attemptig the LP relaxatio. Whe solvig iteger programs, it is importat to idetify these types of problem-depedet restrictios. They ofte occasio large reductios i the computatioal effort.
Table 9. B&B results for warehouse locatio problem Node, k Level, l P k z LB y LB z B y B s Actio 0 0 Ø 790.0 (0, 0.5, 1) M 2 Set y 2 =1 1 1 (+2) 860.0 (0, 1, 0.5) M 3 Set y 3 =3 2 2 (+2, +3) 910.0 (0, 1, 1) 910 (0, 1, 1) Backtrack 3 2 (+2, 3) 932.5 (1, 1, 0) 910 (0, 1, 1) Fathom ad backtrack 4 1 ( 2) 792.0 (0.5, 0,1) 910 (0, 1, 1) 1 Set y 1 =1 5 2 ( 2, +1) 857.5 (1, 0, 0.5) 910 (0, 1, 1) 3 Set y 3 =1 6 3 ( 2, +1, +3) 885.0 (1, 0, 1) 885 (1, 0, 1) Backtrack 7 3 ( 2, +1, 3) Ifeas. 885 (1, 0, 1) Backtrack 8 2 ( 2, 1) Ifeas. 885 (1, 0, 1) Stop Additive Algorithm for the Pure 0 1 Iteger Programmig We ow preset a B&B algorithm that ca be used to solve a 0-1 iteger program without relyig o liear programmig to fid upper bouds. The approach is due to Ego Balas ad is referred to as the additive algorithm. We write the model as Maximize  j =1 c j x j subject to  a ij x j b i, i = 1,,m j =1 x j = 0 or 1, j = 1,, For reasos that will soo become apparet it is assumed that all costraits are of the less tha or equal to type. If a model is ot i this form, the followig trasformatios ca be used to achieve it. If some costrait i is of the greater tha or equal to type, make the followig substitutio.  a ij x j b i fi  a ij x j b i j =1 j =1
If some costrait i is a equality, replace it with the followig iequalities.  a ij x j = b i fi j=1 Ï Â j=1 Ô Ì Ô Ó a ij x j b i  a ij x j b i j=1 It is also assumed that all coefficiets c j i the objective fuctio are opositive. If c j > 0, we replace x j with 1 ^x j, where ^x j is a biary variable. This trasformatio itroduces costats o the left-had side of the costraits that must be moved to the right. Costat terms i the objective fuctio are igored durig the optimizatio but added back whe the solutio is foud. As a illustratio of the variable trasformatio step, cosider the kapsack example below. Maximize z = 5x 1 + 3x 2 + 7x 3 subject to 4x 1 + 2x 2 + 5x 3 8 x j = 0 or 1, j = 1, 2, 3 To obtai egative coefficiets i the objective, each of the variables must be trasformed. The revised problem i the form required by the algorithm is Maximize z = 5^x 1 3^x 2 7^x 3 + 15 subject to 4^x 1 2^x 2 5^x 3 8 11 = 3 ^x j = 0 or 1, j = 1, 2, 3 To use implicit eumeratio as the solutio techique, it is ecessary to defie a relaxatio that will provide bouds ad allow us to test for feasibility.
Boud A very simple relaxatio of a iteger program is oe that requires oly additio to obtai a solutio. If we are at ode k i the search tree with sets S + k, S k ad S 0 k, the problem uder cosideratio ca be writte z k UB = Maximize  jœs 0 k cj x j +  subject to  jœs 0 k jœs + k cj aij x j r i, i = 1,,m (6) x j = 0 or 1, j Œ S 0 k where the secod term i the objective fuctio is a costat -- the cotributio of the variables fixed at 1. Of course, the variables fixed at 0 do ot affect the objective fuctio. To simplify the otatio, we have itroduced r i to represet the right-had side of costrait i; that is, r i = b i  aij, i = 1,,m jœs + k The value r i is the origial right-had-side value less the coefficiets of the variables that are set to 1. Because c j 0, a relaxed solutio x k that maximizes the objective is obtaied by settig all the free variables x j = 0, j Œ S 0 k. The correspodig objective value is computed by summig the coefficiets of the variables fixed to 1; that is, z k UB = S jœs + c j. I additio, if r = k (r 1,...,r m ) 0 implyig that all the costraits are satisfied, x k is optimal to the IP at ode k. I this case, we update the icumbet by puttig z B max{z B, z k UB } ad backtrack. If some of the costraits are ot satisfied whe all free variables are set to zero, eglectig these costraits is a relaxatio of the problem. The objective value obtaied is a upper boud. The solutio to this relaxatio is always iteger, but some of the costraits i (6) may be violated. For the trasformed kapsack example, whe all variables are free at ode 0 the boud obtaied is z 0 UB = 0 (the costat 15 will be igored for the remaider of this sectio). Clearly, this is a upper boud o the
optimal objective value sice all the objective coefficiets are egative. Give r 1 = 3 for the costrait, we see, however, that the solutio is ot feasible. Approximate This procedure is aimed at fidig a feasible solutio at a give ode that ca be used i subsequet fathomig tests. For the additive algorithm, we simply examie the values of r i to see if each is positive. If so, a feasible solutio has bee foud. Variable Fixig For a 0-1 iteger program, various logical tests may be icluded to determie whether a particular ode i the search tree ca be fathomed because it admits o feasible solutios. Agai assume that the eumerative process has progressed to ode k ad cosider costrait i. Let t i be the sum of the egative coefficiets of the free variables. Thus t i =  mi{0, aij } jœs 0 k represets the smallest possible left-had-side value for costrait i ad is always opositive. The right-had-side value r i may be positive, zero, or egative. Whe all the r i are oegative, the solutio to the relaxatio is feasible ad optimal for the ode. If r i is egative for some costrait i ad t i > r i, there is o feasible solutio i the set represeted by the ode. This follows because costrait i caot be satisfied eve if all the free variables with egative structural coefficiets are set to 1 ad the remaiig free variables set to 0. This is oe feasibility test. If a ode fails the test, it is fathomed ad the process backtracks. For example, cosider the costrait  jœs 0 k aij x j = 6x 1 + 5x 2 + 2x 3 3x 4 12 = r i The computatios give t i = 9 > r i = 12 so ode k is fathomed. For the kapsack example at ode 0, r 1 = 3 ad t 1 = 4 2 5 = 11 so the feasibility test is satisfied. This idicates that we must cotiue to eumerate.
Similar reasoig ca be used to determie whe free variables must be fixed to 1 or 0 to assure feasibility. If a ij > 0, j Œ S 0 k ad a ij + t i > r i, x j must be set to 0 for feasibility. If a ij < 0, j Œ S 0 k ad a ij + t i > r i, x j must be set to 1 for feasibility. I this maer, variables may be fixed outside the usual eumeratio process thus reducig the size of the search tree. For the kapsack example, these coditios do ot idicate that ay of the variables ca be fixed. These tests together with the simple relaxatio, ca be icorporated i a implicit eumeratio scheme to solve the pure 0-1 IP. Because oly additio is used at each step i the computatios, the full procedure is called the additive algorithm. Brach The eumeratio procedure requires that a separatio variable be chose at each live ode. Oe method is to choose the variable that most reduces the ifeasibility of the curret solutio. To implemet this idea let R k = {j : j Œ S 0 k ad a ij < 0 for some i such that r i < 0} If there is o i such that r i < 0, the ode is fathomed; otherwise, at least oe variable whose idex is a elemet of R k must equal 1 i ay feasible solutio. Therefore, we ca separate o some x j, j Œ R k, ad the brach to the successor ode correspodig to x j = 1. The followig rule chooses such a j Œ R k i a attempt to move toward feasibility. Defie m I k =  max{0, ri } i=1 to be the ifeasibility of (6). By choosig x j for brachig, the ifeasibility at the successor ode is m I k (j) =  i=1 max{0, ri + a ij } We choose x s to miimize this value; that is,
I k (s) = mi jœr k I k (j) For example, if the costraits at ode k are 6x 1 2x 2 + 2x 3 3 3x 1 4x 2 + x 3 2 7x 1 + 5x 2 5x 3 4 the R k = {1, 2}, I k (1) = 3 ad I k (2) = 2 so x 3 is chose as the separatio variable. Adaptig, oce agai, the depth-first rule for brachig simplifies the represetatio of the search tree. A cosequece of this rule ad of brachig to x k = 1 is that the path vector P k uiquely determies the remaiig eumeratio required. Example 4 (Kapsack Problem) Cosider the trasformed kapsack problem itroduced above. Table 10 describes the solutio obtaied with additive algorithm. At ode 0, the ifeasibility idex I 0 = 3 ad R 0 = {1, 2, 3}. Also, I 0 (1) = 0, I 0 (2) = 1 ad I 0 (3) = 0 implyig that both ^x 1 ad ^x 3 will reduce the ifeasibility to 0 if either is set to 1. We arbitrarily choose ^x 3. At ode 1, the value of r 1 is positive so a feasible solutio has bee obtaied. The algorithm thus backtracks to ode 2 where the variable fixig procedure idicates that ^x 1 should be set to 1. Brachig to ode 3 yields a secod feasible solutio with P 3 = ( 3, +1). Because all the compoets of P 3 are uderlied, the termiatio criterio is met ad the computatios cease. The optimal solutio is ˆ x = (1, 0, 0) with z B = 5. I terms of the origial problem statemet we have x = (0, 1, 1) with z IP = 10.
Table 10. Additive algorithm results for kapsack example Node, k Level, l P k z UB t 1 r 1 z B ˆ x B s Actio 0 0 Ø 0 11 3 M 3 Set ^x 3 =1 1 1 (+3) 7 6 2 7 (0, 0, 1) Backtrack 2 1 ( 3) 0 6 3 7 (0, 0, 1) Fix variable ^x 1 = 1 3 2 ( 3, +1) 5 2 1 5 (1, 0, 0) Stop
13. You are usig Balas s additive algorithm to solve the followig 0-1 IP. Maximize 10x 1 5x 2 7x 3 2x 4 8x 5 11x 6 4x 7 12x 8 subject to 3x 1 x 2 5x 3 3x 4 + 2x 5 + 2x 6 + 8x 7 + 4x 8 6 2x 1 + 3x 2 + 1x 3 10x 4 + 1x 5 4x 6 6x 7 + 2x 8 9 5x 1 4x 2 10x 3 + 2x 4 2x 5 3x 6 5x 7 3x 8 8 3x 1 + 3x 2 + 4x 3 5x 4 + 5x 5 x 6 + 8x 7 4 x 8 4 x j = 0 or 1, for j = 1,,8 a. At ode 0 of the search tree, use the feasibility tests discussed at the ed of Sectio 8.3 (variable fixig) to fix as may variables as possible. Determie the first separatio variable. b. Say you are at a itermediate poit i the eumeratio process. The followig iformatio cocerig the search tree is give: P k = ( 5, +1, +2, 8) ad z B = 99,999 Execute the additive algorithm for as may iteratios as required util backtrackig is idicated. Costruct a table showig all iformatio associated with the odes created i the search tree from the curret poit forward icludig the ode reached after the backtrack operatio. 14. Cosider the 0-1 kapsack problem below. Maximize z = 84x 1 + 48x 2 + 25x 3 + 29x 4 + 128x 5 subject to 49x 1 + 30x 2 + 19x 3 + 29x 4 + 91x 5 130 x j = 0 or 1, j = 1,,5 a. Solve with Balas's algorithm by had showig the implicit eumeratio tree that results. Do ot use a computer. b. Solve by had usig brach ad boud with LP relaxatio. Costruct the search tree ad show all computatios. 20. You are give the followig 0-1 iteger program.
Maximize 10x 1 5x 2 7x 3 2x 4 8x 5 11x 6 4x 7 12x 8 subject to 3x 1 x 2 5x 3 3x 4 + 2x 5 + 2x 6 + 8x 7 + 4x 8 6 2x 1 + 3x 2 + x 3 10x 4 + x 5 4x 6 6x 7 + 2x 8 9 5x 1 4x 2 10x 3 + 2x 4 2x 5 3x 6 5x 7 3x 8 8 3x 1 + 3x 2 + 4x 3 5x 4 + 5x 5 x 6 + 8x 7 4x 8 4 x j = 0 or 1, j = 1,,8 a. For ode 0 of the search tree, use Balas s feasibility tests to fix as may variables as possible. Determie the first separatio variable. b. Say you are at a itermediate poit i the eumeratio process. The followig iformatio cocerig the search tree is give at ode k. S + k = {1, 2}, S k = {5, 8} P k = ( 5, 1, 2, 8), z B = 99,999 Use Balas s algorithm to go as may steps as ecessary util backtrackig is idicated. Show S + k, S k, P k ad z B for all odes created i the tree icludig the ode reached after the backtrack operatio. 21. For the problem Maximize 84x 1 + 48x 2 + 25x 3 + 29x 4 + 128x 5 subject to 49x 1 + 30x 2 + 19x 3 + 29x 4 + 91x 5 130 x j = 0 or 1, j = 1,,5 a. Solve with Balas's algorithm by had showig the complete search tree geerated ad all calculatios. b. Solve by had usig brach ad boud with liear programmig relaxatio at each ode. Show the complete search tree ad all computatios.
31. You are give the followig 0-1 iteger program for which you are usig the additive algorithm to solve. Maximize 10x 1 30x 2 20x 3 20x 4 10x 5 subject to 8x 1 12x 2 x 3 8x 4 2x 5 16 9x 1 7x 2 4x 3 10x 4 x 5 15 6x 1 x 2 8x 3 3x 4 7x 5 9 x j = 0 or 1, j = 1,,5 At the curret iteratio, the solutio sets are S = {2, 4}, S + =. a. Draw the correspodig search tree. Startig from this poit use the feasibility tests to fix as may variables as possible. Commet o the solutio obtaied. b. Cotiue iteratig ad fid the optimum.